Let $$\int \frac{2 - \tan x}{3 + \tan x} dx = \frac{1}{2}(\alpha x + \log_e|\beta \sin x + \gamma \cos x|) + C$$, where $$C$$ is the constant of integration. Then $$\alpha + \frac{\gamma}{\beta}$$ is equal to :

$$\int \frac{2-\tan x}{3+\tan x}dx = \int \frac{2\cos x - \sin x}{3\cos x + \sin x}dx$$.

Write $$2\cos x - \sin x = A(3\cos x + \sin x) + B(-3\sin x + \cos x)$$, where the second part is the derivative of the denominator.

$$2\cos x - \sin x = (3A+B)\cos x + (A-3B)\sin x$$

$$3A + B = 2$$ and $$A - 3B = -1$$.

From second: $$A = 3B - 1$$. Substituting: $$9B - 3 + B = 2 \implies 10B = 5 \implies B = \frac{1}{2}$$.

$$A = \frac{3}{2} - 1 = \frac{1}{2}$$.

$$\int \frac{2\cos x - \sin x}{3\cos x + \sin x}dx = \frac{1}{2}\int dx + \frac{1}{2}\int \frac{-3\sin x + \cos x}{3\cos x + \sin x}dx$$

$$= \frac{1}{2}x + \frac{1}{2}\ln|3\cos x + \sin x| + C$$

Comparing with $$\frac{1}{2}(\alpha x + \ln|\beta\sin x + \gamma\cos x|) + C$$:

$$\alpha = 1$$, $$\beta = 1$$, $$\gamma = 3$$.

$$\alpha + \frac{\gamma}{\beta} = 1 + 3 = 4$$.

The correct answer is Option 2: 4.